Just same old square

If $\alpha = \angle PAB$ and $\beta = \angle PAD$ , then $\cos\alpha \; = \; \frac{1 + X^2 - 9}{2X} \; = \; \frac{X^2-8}{2X} \hspace{2cm} \cos\beta \; = \; \frac{1 + X^2 - 7}{2X} \; = \; \frac{X^2-6}{2X}$ where $X$ is the side of the square. Since $\alpha + \beta = \tfrac12\pi$ , we see that $\begin{aligned} 1 & = \; \cos^2\alpha + \cos^2\beta \\ 4X^2 & = \; (X^2-8)^2 + (X^2-6)^2 \\ X^4 - 16X^2 + 50 & = \; 0 \\ (X^2 - 8)^2 & = \; 14 \end{aligned}$ Since $\alpha$ is acute, we must have $X^2 > 8$ , and hence $X^2 = \boxed{8 + \sqrt{14}}$ .

draw the square on the x-y plane L is the side length

$B=\left ( 0,0 \right ) ; A=\left ( 0,L \right ) ;D=\left ( L,L \right ) ; P=\left ( a,b \right ) \\ \left | \overline{ PB} \right |^2 = a^2+b^2=9\rightarrow (1)\\ \left | \overline{ PA} \right |^2 = a^2+\left ( L-b \right )^2=1\rightarrow (2)\\ \left | \overline{ PD} \right |^2 = \left ( L-a \right )^2+\left ( L-b \right )^2=7\rightarrow (3)\\ (1)-(2)\Rightarrow b=\frac{L^2+8}{2L}\\ (3)-(2)\Rightarrow a=\frac{L^2-6}{2L}$

substitute the value of a and b in (1)

$\frac{\left ( L^2-6 \right )^2+\left (L^2+8 \right )^2}{4L^2}=9\\ L^4-16L^2+50=0\Rightarrow L^2=8\pm \sqrt{14} \\ \color{#D61F06} L=\sqrt{8-\sqrt{14}}\Rightarrow diagonal=\sqrt{2} \cdot \sqrt{8-\sqrt{14}}<3 \\ \color{#D61F06} diagonal<\left | \overline {PA}\right |\Rightarrow 8-\sqrt{14} rejected$

Rotate triangle $\Delta APD$ about $A$ by a right angle in the anticlockwise direction. Then $D$ lands on $B$ . Let $P$ land on some $P'$ . Join $P$ to $P'$ .

Then, $\Delta PAP'$ is an isoceles right-angled triangle right-angled at $A$ , and $PP' = \sqrt{2}$ .

This also means $PBP'$ is a right-angled triangle, right-angled at $P'$ , due to Pythagoras' Theorem, as $PP'^2 + P'B^2 = 7+2=9=PB^2$ .

Thus $\angle AP'B = \angle AP'P + \angle PP'B = 45^{\circ} + 90^{\circ} = 135^{\circ}$

so $AB^2 = P'A^2 + P'B^2 - 2 \cdot P'A \cdot P'B \cdot \cos \angle AP'B = 8+\sqrt{14}$

From point P, draw perpendicular lines from P to AD (=x), and P to AB(= y). We then have the three equations: (1) x^2 + y^2 = 1. (2) (S - y)^2 + x^2 = 7, and (3) (S - x)^2 + y^2 = 9, where S= side of the square. Expanding (2) and (3),and substituting (1), we have the equations: S^2 - 2Sy - 6 = 0 and S^2 - 2Sx - 8 = 0 So 2Sy = S^2 - 6 and 2Sx = S^2 - 8. Multiplying, 4S^2 xy = S^4 - 14S^2 + 48. By subtracting the 2 equations, we have 1 = S(y - x). Squaring, 1 = S^2(y^2 - 2xy + y^2). Usung x^2 + y^2 =1, 1 = S^2 (1 - 2xy) = S^2 - 2S^2 xy. Multiplying by 2, and transposing, 4S^2*xy = 2S^2 - 2.We now have 2 expressions for 4xyS^2. Equating them, we have: 2S^2 - 2 = S^4 - 14S^2 + 48. Transposing, we have S^4 - 16S^2 + 50 = 0, a quadratic equation in S^2, with solution S^2 = [16 +/- sqrt(256 - 200)]/2, or S^2 = 8 + sqrt(14). Ed Gray